CBSE 2024 · Region 1 · Set 2 · Q22 · 2 marks
If $\displaystyle \overrightarrow{\mathrm{a}}$ and $\displaystyle \overrightarrow{\mathrm{b}}$ are two non-zero vectors such that $\displaystyle (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \perp \overrightarrow{\mathrm{a}}$ and $\displaystyle (2 \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \perp \overrightarrow{\mathrm{b}}$, then prove that $\displaystyle |\overrightarrow{\mathrm{b}}|=\sqrt{2}|\overrightarrow{\mathrm{a}}|$.
Marking-scheme solution
$$\begin{align*}
& (\vec{\mathrm{a}}+\vec{\mathrm{b}}) \cdot \vec{\mathrm{a}}=0 \Rightarrow|\vec{\mathrm{a}}|^{2}+\vec{\mathrm{b}} \cdot \vec{\mathrm{a}}=0 \\
& (2 \vec{\mathrm{a}}+\vec{\mathrm{b}}) \cdot \vec{\mathrm{b}}=0 \Rightarrow 2 \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}+|\vec{\mathrm{b}}|^{2}=0 \tag{2}\\
& 2\left(-|\vec{\mathrm{a}}|^{2}\right)+|\vec{\mathrm{b}}|^{2}=0 \text { \{Using (1) and (2)\}} \\
& |\vec{\mathrm{b}}|^{2}=2|\vec{\mathrm{a}}|^{2} \Rightarrow|\vec{\mathrm{b}}|=\sqrt{2}|\vec{\mathrm{a}}|
\end{align*}
$$
Vector AlgebraProduct of Two VectorsApplyvery_short_answermedium
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